Class 9 Physics Chapter 8 numerical problems are according to the new syllabus of the Punjab Board. Chapter 8 is related to the Thermal Properties of Matter in which you will study the working principles of how matter is converted into different substances using multiple phenomenons.
Class 9 Physics Chapter 8 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
F = 1.8C + 32
T(K) = C + 273
Δ L = α Δ T
V = Vo(1 + βΔ T)
Δ Q = mc Δ T
P = W / t
P = Q / t
ΔQf = m Hf
Q = mL
Numerical Problems Chapter 8 – Thermal Properties of Matter
8.1 Temperature of water in a beaker is 50°C, what is its value in the Fahrenheit scale? (122°F)
Solution:
Temperature in Celsius scale = C = 50°C
Temperature in Fahrenheit scale = F = ?
F = 1.8C + 32
F = 1.8 + 32
F = 90 + 32
F = 122°F
8.2 Normal human body temperature is 98.6°F, convert it into Celsius scale and kelvin scale. (37°C,310K)
Solution:
Temperature in Fahrenheit scale = 98.6°F
Temperature in Celsius scale = ?
Temperature in Kelvin scale = ?
F = 1.8C + 32
1.8C = F – 32
1.8C = 98.6 -32
1.8C = 66.6
C = = 37°C
T(K) = C + 273
T(K) = 37 + 273
T(K) = 310K
8.3 Calculate the increase in the length of an aluminium bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminium is 2.5. (0.1cm)
Solution:
Original length of rod = 2m
Initial temperature = 0°C = 0 + 273 = 273 K
Final temperature = T = 20°C = 20 + 273 = 293 K
Change in temperature = Δ T = T – = 293 – 273 = 20 K
Coefficient of linear expansion of aluminum = α = 2.5
Increase in volume Δ L = ?
Δ L = α Δ T
Δ L = 2.5
Δ L = 100
Δ L = 0.001 m = 0.001 100 = 0.1cm
8.4 A balloon contains 1.2 air at 15°C. Find its volume at 40°C. Thermal coefficient of volume expansion of air is 3.67 . (1.3m3)
Solution: Original volume = 1.2
Initial temperature = = 15°C = 15 + 273 = 288 K
Final temperature = T = 40°C = 40 + 273 = 313 K
Change in temperature = Δ T = T – To = 313 – 288 = 25 K
Coefficient of volume expansion of air β = 3.67
Volume = V =?
V = Vo(1 + βΔ T)
V = 1.2 (1 + 3.67 x 10-2 x 25)
V = 1.2(1 + 0.09175)
V = 1.2(1.09175)
V = 1.3m3
8.5 How much heat is required to increase the temperature of 0.5 kg of water from 10°C to 65°C? (115500 J)
Solution:
Mass of water = m = 0.5 kg
Initial temperature = T1 = 10°C = 10 + 273 = 283 K
Final temperature = T2 = 65°C = 65 + 273 = 338 K
Change in temperature = Δ T = T2 – T1 = 338 – 283 = 55 K
Heat = Δ Q =?
Δ Q = mc Δ T
Δ Q = 0.5 x 2400 x 55
Δ Q = 115500J
8.6 An electric heater supplies heat at the rate of 1000J per second. How much time is required to raise the temperature of 200 g of water from 20°C to 90°C? (58.8 s)
Solution:
Power = P = 1000 Js-1
Mass of water = m = 200 g = 200/1000 = 0.2 kg
Initial temperature = T2= 20°C = 20 + 273 = 293 K
Final temperature = T1 = 90°C = 90 + 273 = 363 K
Change in temperature = Δ T = T2 – T1 = 363 – 293 = 70 K
Specific heat of water = c = 4200 Jkg-1 K-1
Time = t =?
P = W / t
Or P = Q / t
Or P t = Q
Or P t = mc Δ T
Or t= mc Δ T / P
t = 0.2 x 4200 x 70 / 1000
t = 58.8 s
8.7 How much ice will melt by 5000 J of heat? Latent heat of fusion of ice = 336000 J. (149 g )
Solution:
Amount of heat required to melt ice = ΔQf = 50000J
Latent heat of fusion of ice = Hf = 336000 J
Amount of ice = m =?
ΔQf = m Hf
Or m = ΔQf / Hf
m = 50000 / 336000 = 0.1488 kg
m = 0.1488 x 1000 g
m = 1488 x 1000 / 1000
m = 148.8 g ≈ 149 g
8.8 Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C. (39900 J)
(Note: Specific heat of ice is 2100 J, and the specific heat of water is 4200 J Latent heat of the fusion of ice is 336000 J .
Solution:
Mass of ice = m = 100 g = = 0.1 kg
Specific heat of ice = c1 = 2100 J
Latent heat of fusion of ice = L = 336000 J
Specific heat of water = c = 4200 J
Quantity of heat required = Q =?
Case I:
Heat gained by ice from -10°C to 0°C
Q1 = mc Δ T
Q1 = 0.1 2100 10 = 2100 J
Case II:
Heat required for ice to melt = Q2 = mL
Q2 = 0.1 336000
Q2 = 33600 J
Case III:
Heat is required to raise the temperature of water from 0°C to 10°C
Q3= mc Δ T
Q3 = 0.1 x 4200 x 10 = 4200 J
Total heat required = Q = Q1 + Q2 + Q3
Q = 2100 + 33600 + 4200
Q = 39900 J
8.9 How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is 2.26 . (2.26 J)
Solution:
Mass of water = m = 100 g = 100 / 1000 = 0.1 kg
Latent heat of vaporization of water = Hv = 2.26 x 106 jkg-1
Heat required = ΔQv =?
ΔQv = mHv
ΔQv = 0.1 x 2.26 x 106 = 0.226 x 106 = 226 x 106 / 1000
ΔQv = 2.26 x 10-1 x 106 = 2.26 x 105 J
8.10 Find the temperature of the water after passing 5 g of steam at 100°C through 500 g of water at 10°C. (16.2°C)
(Note: Specific heat of water is 4200 Jkg-1 K-1, Latent heat of vaporization of water is 2.26 x 106 jkg-1 K-1).
Solution:
Mass of stream = m1 = 5 g = 5 / 1000 kg = 0.005 kg
Temperature of stream = T1 = 100°C
Mass of water = m2 = 0.5 kg
Temperature of water = T2 = 10°C
Final temperature = T3 = ?
Case I:
Latent heat lost by stream = Q1 = mL
Q1 = 0.005 x 2.26 x 106 = 11.3 x 103 = 11300 J
Case II:
Heat lost by stream to attain final temperature = Q2 = m1c Δ T
Q2 = 0.005 x 4200 x (100 – T3)
Q2 = 21 (100 – T3)
Case III:
Heat gained by water Q3 = m2c Δ T
Q3 = 0.5 x 4200 x (T3 – 10)
Q3 = 2100 (T3 – 10)
According to the law of heat exchange.
Heat lost by stream = heat gained by the water
Q1 + Q2 = Q3
11300 + 21 (100 – T3) = 2100 (T3 – 10)
11300 + 2100 – 21T3 = 2100T3 – 21000
13400 + 21000 = 2100T3 + 21T3
34400 = 2121T3
T3 = 34400 / 2121
T3 = 16.2 °C
Class 9 Physics Full Book Numerical Problems with Solution
